consider

4

1 2 2 3

then prefix sum=>1 3 5 8

suffix sum=>3 5 7 8

possible pairs are {1,7} {3,5} {3,5}

1 3 3

series 1=>1 2 0 5

arranging(1 3 3)=> 3 3 1

series 2=>3 0 -2 7

arranging(1 3 3)=> 3 1 3

series 3=>3 -2 2 5

Next possible pair=>1 5 5

similarly i will do like above and get 3 possible arrangements

Next possible pair=>1 3 5

similarly i will do like above and get 6 possible arrangements

Next possible pair=>1 5 3

similarly i will do like above and get 6 possible arrangements

Next possible pair=>7 5 5

similarly i will do like above and get 3 possible arrangements

Next possible pair=>7 3 5

similarly i will do like above and get 6 possible arrangements

Next possible pair=>7 3 3

similarly i will do like above and get 3 possible arrangements

Next possible pair=>7 5 3

similarly i will do like above and get 6 possible arrangements

ANS=>36

IS THIS CORRECT??

It’s 0. You are given the prefix and suffix sums. You can’t take the prefix sum of the input.

Also 1 5 3 is arranged 6 times and so is 1 3 5. Which is overcounting. Same can be said for 7 3 5 and 7 5 3. So the answer is 24.

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**THANKS !!!**

consider

4

1 3 5 8 3 5 7 8

possible pairs are {1,7} {3,5} {3,5}

1 3 3

series 1=>1 2 0 5

arranging(1 3 3)=> 3 3 1

series 2=>3 0 -2 7

arranging(1 3 3)=> 3 1 3

series 3=>3 -2 2 5

Next possible pair=>1 5 5

similarly i will do like above and get 3 possible arrangements

Next possible pair=>1 3 5

similarly i will do like above and get 6 possible arrangements

Next possible pair=>7 5 5

similarly i will do like above and get 3 possible arrangements

Next possible pair=>7 3 5

similarly i will do like above and get 6 possible arrangements

Next possible pair=>7 3 3

similarly i will do like above and get 3 possible arrangements

ANS=>24

(3! / 2!)*2^3=24

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Also you can use`$ (3!/2!) *2^3$`

to make it look neater. It will look like this

(3!/2!) *2^3

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